# Peskin Schroeder page 30 eq 2.54

1. May 9, 2013

### silverwhale

Hello Everybody,

I am trying to get the second line of 2.54 from the last line; I want to get:

$$\int \frac{d^3p}{{2 \pi}^3} \{ \frac{1}{2 E_\vec{p}} e^{-ip \cdot (x-y) }|_{p^0 = E_\vec{p}} + \frac{1}{-2 E_\vec{p}} e^{-ip \cdot (x-y) }|_{p^0 = -E_\vec{p}} \},$$

from

$$\int \frac{d^3p}{{2 \pi}^3} \int \frac{dp^0}{2 \pi i} \frac{-1}{p^2 - m^2}e^{-ip \cdot (x-y) }.$$

Now, my question is: why are the residues of the integrand of the p^0 integral at the two poles ±E_p given by:

$$Res_{p^0=E_\vec{p}} \{ \frac{-1}{(p^0)^2 - (E_\vec{p})^2}e^{-ip^0 \cdot (x_0-y_0) } \}= \frac{1}{2 E_\vec{p}} e^{-i E_\vec{p} \cdot (x^0-y^0) },$$ and
$$Res_{p^0=-E_\vec{p}} \{ \frac{-1}{(p^0)^2 - (E_\vec{p})^2}e^{-ip^0 \cdot (x_0-y_0) } \}= \frac{1}{-2 E_\vec{p}} e^{i E_\vec{p} \cdot (x^0-y^0) }.$$

2. May 9, 2013

### vanhees71

I hope that this is not what's written in Peskin Schroeder, although this book has to be read with great care, because there are a lot of subtle typos in it, where the reader gets confused a lot (at least I am, when looking at this book).

I guess, what you want to calculate is the space-time representation of the time-ordered propagator (in the vacuum theory, i.e., at 0 temperature and density, that's identical with the Feynman propagator) of the Klein-Gordon field. It is given as a Fourier integral:
$$D(x)=\int_{\mathbb{R}^4} \frac{\mathrm{d}^4 p}{(2 \pi)^4} \frac{1}{p^2-m^2+\mathrm{i} 0^+} \exp(-\mathrm{i} p \cdot x).$$
Here, I use the west-coast convention for the metric, i.e., $p \cdot x=p^0 x^0-\vec{p} \cdot \vec{x}$. The little imaginary part in the denominator is VERY important, and must never ever be omitted anywhere in any QFT textbook. I hope that at least this wasn't done by Peskin and Schröder in their book, although I remember that I've seen such a sin in some QFT textbook!

Anyway. The trick now is to perform the $p^0$ integral, using the residue theorem. To that end we write the denominator a bit
$$p^2-m^2+\mathrm{i} 0^+=(p^0)^2-\omega^2 + \mathrm{i} 0^+,$$
where $\omega=\sqrt{\vec{p}^2+m^2}$ denotes the on-shell energy of the particle.

Now to apply the theorem of residues you have to close the integration contour along the real axis. This can be done by a semi-circle in the upper (lower) half-plane if $x^0<0$ ($x^0>0$), because then due to the exponential function in the Fourier integral the contribution of this semi-circle is 0.

For $x^0<0$ you have to find the residues in the upper half-plane. Now our denominator can be written as
$$(p^0)^2-\omega^2+\mathrm{i} 0^+=(p^0-\omega+\mathrm{i} 0^+)(p^0+\omega-\mathrm{i} 0^+),$$
because when you resolve the product, you get
$$(p^0-\omega+\mathrm{i} 0^+)(p^0+\omega - \mathrm{i} 0^+)=(p^0)^2-(\omega-\mathrm{i} 0^+)^2 = (p^0)^2-(\omega^2-2 \mathrm{i} \omega 0^+)=(p^0)^2-\omega^2+\mathrm{i} 0^+.$$
This means that you have two poles in your integrand at
$$p^0=\pm \omega \mp \mathrm{i} 0^+.$$

For $x^0<0$ you need the pole in the upper plane, which is $p^0=-\omega+\mathrm{i} 0^+$. Since this is a simple pole, the residue is calculated from the integrand by
$$\mathrm{res}_{p^0=-\omega+\mathrm{i} 0^+}=\lim_{p^0 \rightarrow -\omega + \mathrm{i} 0^+} (p^0+\omega+\mathrm{i} 0^+) \frac{\exp(-\mathrm{i} p^0 x^0)}{(p^0)^2-\omega^2+\mathrm{i} 0^+} = -\frac{\exp(-\mathrm{i} \omega x^0)}{2 \omega}.$$
So you get
$$D(x)=-\mathrm{i} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{\exp(-\mathrm{i} \omega x^0)}{2 \omega} \exp(\mathrm{i} \vec{x} \cdot \vec{p}), \quad x^0<0.$$
In the same way you find for $x^0>0$ (noting that you have to close the contour in the lower plane, i.e., integrating clockwise, leading to an additional sign in the residue theorem):
$$D(x)=-\mathrm{i} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{\exp(+\mathrm{i} \omega x^0)}{2 \omega} \exp(\mathrm{i} \vec{x} \cdot \vec{p}), \quad x^0>0.$$
Here you can substitute $\vec{p} \rightarrow -\vec{p}$, and you can write
$$D(x)=-\mathrm{i} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{1}{2\omega} \left [\Theta(-x^0) \exp(-\mathrm{i} p \cdot x)+\Theta(x^0) \exp(+\mathrm{i} p \cdot x) \right]_{p^0=\omega)}.$$

3. May 9, 2013

### andrien

The way the integration is performed is by giving mass a negative imaginary part.Otherwise it is clear that there is ambiguity regarding the poles which lie on real axis.this rule often works and remove this ambiguity.

4. May 9, 2013

### silverwhale

Many thanks vanhees71! Wonderful explanation!! Got it from first reading!

And well yes, P&S do omit the +-i epsilon term in the denominator of the integrand.

Thanks andrien for the clarification regarding the imaginary part!

5. May 10, 2013

### vanhees71

The emphasis must ly on "often", i.e., it works sometimes but not always. You have to know, what you like to calculation. In vacuum QFT, you indeed usually are after the S-matrix, and you usually have to evaluate it in perturbation theory (or more sophisticated resummations thereof), and there you indeed need the time-ordered propagator, because you get the S-matrix from the time-evlolution operator for the states in the interaction picture, and this involves a time-ordering operator from the integration of its equation of motion.

In many-body theory, however, you usually have to evaluate expectation values of observables, and here the issue becomes more complicated. In thermal equilibrium you can use the imaginary-time formualism, where you use the Matsubara propagator. In both thermal equilibrium or nonequilibrium you can use the real-time formalism. There it turns out that you do not need the time-ordered propagator only but in addition also the fixed-ordered Wigthman functions (and the anti-time-ordered propagator) as well. To evaluate the reaction of a system to some external disturbance (e.g., an electromagnetic field in interaction with the matter discribed by many-body QFT) using linear-response theory, you need the retarded propagator.

So you always have to carefully derive, which propagator you have to use!

6. May 11, 2013

### silverwhale

Although I carefully read the text yesterday I started to do the calculations today. And I already have some questions!

In vanhees calculations, there is the following line:

$$\mathrm{res}_{p^0=-\omega+\mathrm{i} 0^+}=\lim_{p^0 \rightarrow -\omega + \mathrm{i} 0^+} (p^0+\omega+\mathrm{i} 0^+) \frac{\exp(-\mathrm{i} p^0 x^0)}{(p^0)^2-\omega^2+\mathrm{i} 0^+} = -\frac{\exp(-\mathrm{i} \omega x^0)}{2 \omega}.$$

Now I already have a question here. Shouldn't the "last" RHS be:

$$= -\frac{\exp(+\mathrm{i} \omega x^0)}{2 \omega} ?$$

What am I missing here?

I also returned to Peskins calculations (eq 2.54) and successfully determined the residues. Now what spoils my calculation is a minus sign in front of the one in the numerator of the integrand of the p^0 integral in the last line:

$$\int \frac{d^3p}{{2 \pi}^3} \int \frac{dp^0}{2 \pi i} \frac{-1}{p^2 - m^2}e^{-ip \cdot (x-y) }.$$

Without the minus sign I do get the previous line

$$\int \frac{d^3p}{{2 \pi}^3} \{ \frac{1}{2 E_\vec{p}} e^{-ip \cdot (x-y) }|_{p^0 = E_\vec{p}} + \frac{1}{-2 E_\vec{p}} e^{-ip \cdot (x-y) }|_{p^0 = -E_\vec{p}} \}.$$

Is it because I am integrating counter clockwise as I am closing the integral below? So that I get an overall minus sign which cancels the existing one?

7. May 11, 2013

### vanhees71

You are right! There's a mistake in my previous posting concerning the signs. Unfortunately I cannot edit my own posting anymore. So here are the correct results:

$$\mathrm{res}_{p^0=-\omega+\mathrm{i} 0^+}=\lim_{p^0 \rightarrow -\omega + \mathrm{i} 0^+} (p^0+\omega+\mathrm{i} 0^+) \frac{\exp(-\mathrm{i} p^0 x^0)}{(p^0)^2-\omega^2+\mathrm{i} 0^+} = -\frac{\exp(+\mathrm{i} \omega x^0)}{2 \omega}.$$

Accordingly the correct representation of the time-ordered vacuum propagator for the free scalar field reads
$$D(x)=-\mathrm{i} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{1}{2\omega} \left [\Theta(-x^0) \exp(+\mathrm{i} p \cdot x)+\Theta(x^0) \exp(-\mathrm{i} p \cdot x) \right]_{p^0=\omega}.$$

I've my copy of Peskin-Schroeder here now, and comparing this with his, it's correct. Note, however that he has a somewhat different notation. What I call $D$ he calls $-\mathrm{i} D_F$ (for Feynman propagator, which is correct since in the vacuum the time ordered propagator is identical with the Feynman operator). The question, where to put the factor $\mathrm{i}$ is purely conventional, of course. His $D$ is the Wightman function (fixed ordered vacuum-correlation function of the fields).

Concerning the rest of your questions: This part of Peskin-Schroeder's calculation is at least misleading (in my opinion it's even wrong), because the integral along the real axis without the $\mathrm{i} 0^+$ in the denominator doesn't make sense. Peskin-Schroeder is repairing this in the same section by carefully analyzing the various possibilities to choose the integration path in complex $p^0$ plane to circumvent the poles on the real axis. This leads to different propagators of the Klein-Gordon propagator which, of course, is not uniquely defined by its equation of motion (I'm using my convention concerning factors $\pm \mathrm{i}$)
$$-(\Box+m^2) D(x)=\delta^{(4)}(x).$$
The reason is simple! You can take any solution of this equation, e.g., the retarded propagator, which you would use for a source emitting waves in classical physics. This function is uniquely defined by the demand that
$$D_{\text{ret}}(x)=0 \quad \text{for} \quad x^0<0.$$
Then any other function
$$D(x)=D_{\text{ret}}(x)+\Delta(x),$$
where $\Delta$ is a solution of the homogeneous KG equation,
$$-(\Box+m^2) \Delta(x)=0.$$
So there is not only one propagator! You have, e.g., also the advanced propagator or the time-ordered (Feynman) one or the anti-time-ordered one. It depends on the context, which one to use.

You get these different propagators also in the way Peskin-Schroeder explains it, i.e., by choosing different paths in the $p^0$ integral or by regularizing the integrand with (small) imaginary parts in the denominator (as is also explained by Peskin and Schroeder).

Finally, when using the theorem of residues, you have to close the integration path by large semi circles (letting the radius to $\infty$), but this semi circle must not contribute to the integral and particularly it must not make the integral divergent. Thus you have to choose the semi circle always such that the exponential in the Fourier integral is damped (rather than exponentially raising) at infinity. Thus you have to close the path in the upper plane if $x^0<0$ and in the lower plane, when $x^0>0$. Then it depends on the choice of paths close to the real axis, i.e., how you circumvent the poles of the integrand or on the sign when regularizing the integrand with (small) imaginary parts in the denominator. So, only after choosing the exact type of propagator you want to calculate, you can give a precise meaning to that integrals and have a unique way to calculate it with the theorem of residues.

Due to this confusion, I do not recommend Peskin-Schroeder anymore as a textbook to start QFT. I find Ryder very helpful for that purpose. Later, when a bit familiar with the subject, you should read Weinberg's marvelous books. These are the best QFT texts I'm aware of.

Last edited: May 11, 2013
8. May 11, 2013

### silverwhale

I do have Ryder, got it from the library last week!
And Colemans QFT script! Wonderful script, really plain wonderful!
Well gotta change the books then!! :)

Thanks again vanhees!!!

9. May 11, 2013

### silverwhale

I am redoing the calculations right now I'll report later! :)

10. May 11, 2013

### silverwhale

Basically I got the result:

$$D(x)=-\mathrm{i} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{1}{2\omega} \left [\Theta(-x^0) \exp(+\mathrm{i} p \cdot x)+\Theta(x^0) \exp(-\mathrm{i} p \cdot x) \right]_{p^0=\omega}.$$

But I do get the result only when adding a minus sign to the residue of f at the pole p^0 = omega - i0+.
So I basically say that the p^0 integral is equal to (-) 2 pi i Res(f).

So I want to ask, is this what you mean by:"noting that you have to close the contour in the lower plane, i.e., integrating clockwise, leading to an additional sign in the residue theorem"?

Isn't it very sloppily the prescription of adding that minus sign when the contour is clockwise?

The rest of the text mentioning the conventions you took and how to get the differents Greens functions is clear to me! (I hope!! :> )

11. May 11, 2013

### vanhees71

Why should this be sloppy? The residuum theorem is stated for running counter-clockwise, i.e., for a holomorphic function you have for each point $z$ in the interior of a closed counter-clockwise oriented curve
$$\int_{\gamma} \mathrm{d z'} \frac{f(z')}{z-z'}=2 \pi \mathrm{i} f(z).$$
If $\gamma$ is clockwise oriented, you get an additional - sign on the right-hand side of this equation, because then $-\gamma$ is counter-clockwise oriented and you have the integral-operator relation
$$\int_{\gamma} \mathrm{d} z' = -\int_{-\gamma} \mathrm{d} z'.$$

12. May 11, 2013

### vanhees71

Yes, Coleman is great. I don't know his QFT book but "Aspects of Symmetry", which is a collection of lectures he has given at the Erice schools. Its marvelous.

There are also online movies of his QFT lectures. The link has been posted some time ago here in Physics Forums:

I just realized that there is also a link to a a transcript of the lecture notes on the ArXiv:

http://arxiv.org/abs/1110.5013

13. May 11, 2013

### silverwhale

Yeah those are the lecture notes I mean!

14. May 12, 2013

### silverwhale

Dear vanhees,

When again going through the text I found that you wrote the following quote:"Peskin-Schroeder is repairing this in the same section by carefully analyzing the various possibilities to choose the integration path in complex p0 plane to circumvent the poles on the real axis."

Now, I returned to P&S (please forgive me!) just to check this. And what I found was that (what he writes), there are 4 contours for performing the integral. Now my question is, are there really 4? I hink there should be 8 in total!

1- Integrating without the poles, 2 ways
2- Integrating with one pole included, 2x2 ways
3- Integrating with both poles included, 2 ways.

Am I wrong?
I felt that the courses of complex analysis I had where maybe incomplete, or I am missing practice, or P&S is really far too messy..

15. May 12, 2013

### vanhees71

As far as I can see, there are 4 ways to run around the poles: you can either use a little semicircle in the upper plane or the lower for each of both poles.

16. May 12, 2013

### silverwhale

Then integration without the poles and with both poles included in the contour is ruled out then.

17. May 12, 2013

### silverwhale

Well iny any case adding the i0+ term makes the whole thing easy to grasp and to calculate..

18. May 12, 2013

### vanhees71

The point is that you have to close the contour in the upper for $t<0$ or lower half plane for $t>0$. E.g. for the retarded Green's function you have to choose the contour such that for $t<0$ no pole is within the contour, i.e., there must be no pole in the upper half-plane. With the little imaginary part this is written as
$$\tilde{D}_{\text{ret}}(p)=\frac{1}{(p^0+\mathrm{i} 0^+)^2-\omega^2}.$$
Then, of course both poles are within the contour enclosing the lower half-plane for $t>0$.